Question: Simplify and expand the following expression: $ \dfrac{4}{x - 4}- \dfrac{4}{x + 2}- \dfrac{1}{x^2 - 2x - 8} $
First find a common denominator by finding the least common multiple of the denominators. Try factoring the denominators. We can factor the quadratic in the third term: $ \dfrac{1}{x^2 - 2x - 8} = \dfrac{1}{(x - 4)(x + 2)}$ Now we have: $ \dfrac{4}{x - 4}- \dfrac{4}{x + 2}- \dfrac{1}{(x - 4)(x + 2)} $ The least common multiple of the denominators is: $ (x - 4)(x + 2)$ In order to get the first term over $(x - 4)(x + 2)$ , multiply by $\dfrac{x + 2}{x + 2}$ $ \dfrac{4}{x - 4} \times \dfrac{x + 2}{x + 2} = \dfrac{4(x + 2)}{(x - 4)(x + 2)} $ In order to get the second term over $(x - 4)(x + 2)$ , multiply by $\dfrac{x - 4}{x - 4}$ $ \dfrac{4}{x + 2} \times \dfrac{x - 4}{x - 4} = \dfrac{4(x - 4)}{(x - 4)(x + 2)} $ Now we have: $ \dfrac{4(x + 2)}{(x - 4)(x + 2)} - \dfrac{4(x - 4)}{(x - 4)(x + 2)} - \dfrac{1}{(x - 4)(x + 2)} $ $ = \dfrac{ 4(x + 2) - 4(x - 4) - 1} {(x - 4)(x + 2)} $ Expand: $ = \dfrac{4x + 8 - 4x + 16 - 1}{x^2 - 2x - 8} $ $ = \dfrac{23}{x^2 - 2x - 8}$